Part A Here, the velocity is at an angle of 45 degrees to the x-axis toward the z-axis. The velocity is then given in component form as.

Transcription

1 Electrodynaics Chapter Andrew Robertson Here we are given a proton oving in a agnetic eld ~ B 0:5^{ T at a speed of v :0 0 7 /s in the directions given in the gures. Part A Here, the velocity is at an angle of 45 degrees to the x-axis toward the z-axis. The velocity is then given in coponent for as ~v v cos (45)^{ + v sin (45) ^k p2 0 ^{ 7 + p2 0 7 ^k The agnetic force on this particle is coputed via a cross product as follows: ~F ~v B ~ ^{ ^ ^k p p : p 2 07 ^ Reebering that the charge on a proton is : C, we get a force of ~F 2 p 2 07 ^ N 5: ^ N Part B Here, the velocity is in the direction of the negative x-axis. The velocity is then given in coponent for as ~v v^{ :0 0 7^{ /s The agnetic force on this particle is coputed via a cross product as follows: ~F ~v B ~ ^{ ^ ^k : :5 0 0 ~0 Exactly as expected; A agnetic eld exerts no force on a charged particle oving in the direction (or exact opposite direction) of the eld lines We now wish to know the agnetic eld strength and direction that will levitate the 2.0g wire in the gure. This wire is in a 0c long region of space wherein the agnetic eld is applied. The agnetic force F ~ B ust exactly cancel the weight W ~ of the wire in order to levitate it. That eans the force ust be pointing upward while the current is going fro right to left. Using the right-hand rule, we know that this eans the ~B eld ust be pointing out of the page. We still need to know how strong the eld ust be to copletely cancel out gravity s in uence. We know that the strength of the agnetic force ust atch that of the weight. FB ~ W ~ g (0:002 kg) 9: /s 2 0:096 N We also know that the for of the the agnetic force on a wire carrying current perpendicular to the agnetic eld. ~ FB ILB (:5 A) (0: ) B 0:5B Setting this eual to g and solving for B gives us B g IL 0:096 0:5 0:3 T

2 32.76 Here we are given a nonunifor agnetic eld acting on a loop of current carrying wire (which lies perpendicular to the page). We can tell fro the right hand rule that the agnetic force d F ~ B on any in nitesial segent of the ring will point upward at an angle of 90 to the vertical axis. See the diagra at the end of the docuent. The proble infors us that the net force will be upward on the vertical axis (because of the cylindrical syetry of the proble). As such, we need only concern ourselves with the upward coponent of d F ~ B. That coponent is given by df up cos (90 ) d F ~ B cos (90 ) IB dl where dl is an in nitesial length of the ring. 0 to the circuference of the ring 2R. Part B ~ F net Fup Z 2R 0 We can now get the total force by integrating along dl fro Z 2R cos (90 ) IB dl cos (90 ) IB 0 dl 2R cos (90 If R 0:02, I 0:5 A, B T, and 20, we get the following for the net force. F ~ net 2R cos (90 ) IB 2 (0:02) cos (90 20) (0:5) :0042 N ) IB 32.0 Here we wish to nd the agnetic eld strength that will cancel the torue exerted by a 50g weight hanging on a current carrying loop 2.5 c fro the axis of rotation. If W is the weight of the block and r 2:5 is the distance fro the axis of rotation, we know that the torue exerted by the block has a agnitude of b W r sin (90) gr sin (90) (0:05) (9:) (0:025) () 0:022 N Siilarly, the agnetic torue on a loop in a agnetic eld is eual to the cross product of its agnetic oent and the eld B. ~ Since the angle between ~B and the ~ NI A ~ (where A ~ is the loop s noral area vector) is 90 degrees, we have j~ B j ~ B ~ B sin (90) NIAB sin (90) (0) (2:0) (0:05 0:) B 0:B Setting these two torue agnitude euals to each other, we have B gr NIA 0:022 0:22 T 0: 32.4 Here we need only apply the forula for the agnitude of the agnetic torue on a loop of current carrying wire. j~ B j ~ B ~ B sin (30) IAB sin (30) A 0: (:2 T) sin (30) 7:5 0 4 N 2

3 32.73 We are given a long current carrying wire suspended by threads and de ected fro the vertical by 0 because of the presence of a agnetic eld. We wish to nd the strength and direction of that eld. The right hand rule iediatel gives us that the direction ust be straight downward. Now to copute it s strength, we have the following free body diagra. We shall look down the length of the wire fro one end. Notice that if the strings are de ected by 0, then the tension ust be applied to the wire at an angle of Fro the diagra at the end of the docuent, we can see that the forces ust cancel in the vertical direction. T sin (0) g 0 and in the horizontal direction. Rearranging, we have Dividing the top euation by the botto, we have or T cos (0) F B 0 g T sin (0) F B T cos (0) g F B tan (0) F B g tan (0) To get the agnetic eld, we shall ake the substitutions F B ILB and g Lg where 0:05 kg/ is the linear ass density and F B ILB because ~ B is perpendicular to the wire. Subsituting these into the euation above, we have Solving for B, we have B ILB Lg tan (0) g (0:05) (9:) I tan (0) 0 tan (0) 0:006 T We are now given an electron entering a region with a unifor 30 T agnetic eld in the positive z direction. The electron has a speed v 5:0 0 6 /s directed at an angle 30 above the x-y plane. We want the radius r and the pitch p of the spiral trajectory the electron will follow. We know that the radius of cyclotron orbit is given by r cyc v? B where v? is the agnitude of the velocity coponent that is perpendicular to the agnetic eld. case, this coponent is eual to the coponent of ~v in the xy plane. v? vx 2 + vy 2 p p 3 v 2 vz 2 v 2 v 2 sin 2 (30) v sin 2 (30) v 2 4:33 06 /s For the radius, this gives us r cyc ev? eb 9: 0 3 kg 4: /s (:6 0 9 C) ( :2 0 4 T) 3 In our

4 Next, we realize that the pitch p is eual to how far upward in the z direction the electron goes in the tie it takes to coplete one cyclotron orbit. That tie, called the period T, is the reciprocal of the cyclotron freuency f given in euation p v z T v sin (30) f v sin (30) 2 e B 0: Siply applying the right-hand rule to both of these situations indicates the agnetic force on any straight side of the loop is opposed by the agnetic force on the opposite side. Because that force is eual to F B ILB, which is the sae for each side, we can be sure that the net force on both of these loops is zero. Siilarly, the torue on each loop is ~ ~ B ~ However, ~ and B ~ are either parallel or antiparallel to each other for both loops, hence their cross products vanish. Thus, there is no torue on these loops either. part B For loop, we see that ~ is parallel to B. ~ However, ~ is antiparallel to B ~ for loop 2. So when loop 2 is perturbed slightly fro eulibriu by rotating its oent ~ slightly toward the vector B, ~ the torue exerted will drag ~ further toward B ~ and further away fro euilibriu uch like gravity pulls a rolling ball further fro its euilibriu point at the top of a hill. Siilarly for loop, if perturbed slightly by rotating its oent ~u slightly away fro B, ~ the agnetic torue exerted will try to pull ~u back toward B ~ and toward euilibriu. This is like how gravity pulls a rolling ball back into a well or basin when it tries to escape Here we are given a ass spectroeter. The accelerating potential V iparts a kinetic energy eual to V to a charge particle before it enters the agnetic eld. Solving for v, we have KE 2 v2 V v r 2V It will then undergo a cyclotron orbit of radius d 2 r cyc v B r 2V B Solving for V gives us V 2 Brcyc Bd (Bd)2 2 Thus, we ay easily calcultate the potential di erence V necessary for detecting N + 2 V (0:2 0:0)2 (0:2 0:0)2 5 : C! 2:0062 (: :96 V kg) 4

5 for O + 2 V (0:2 0:0)2 (0:2 0:0)2 7 : C! 3:99 (: :0 V kg) and lastly for CO + V (0:2 0:0)2 (0:2 0:0)2 5 : C! 27:9949 (: :62 V kg) 5

6 N 90 θ F B θ B Proble Proble

7 T 0 F B W 7